To solve the system of equations using the elimination method, we start with the given equations:
[tex]\[
\begin{array}{l}
6x + 2y = 8 \quad (1) \\
12x + y = 22 \quad (2)
\end{array}
\][/tex]
### Step 1: Make the coefficients of [tex]\( y \)[/tex] the same
We can achieve this by manipulating the given equations. One effective approach is to make the coefficient of [tex]\( y \)[/tex] the same in both equations:
- Multiply the first equation by 1 (to keep it the same):
[tex]\[
6x + 2y = 8 \quad (3)
\][/tex]
- Multiply the second equation by 2:
[tex]\[
2 \cdot (12x + y) = 2 \cdot 22 \implies 24x + 2y = 44 \quad (4)
\][/tex]
So now we have the system of equations:
[tex]\[
\begin{array}{l}
6x + 2y = 8 \quad (3) \\
24x + 2y = 44 \quad (4)
\end{array}
\][/tex]
### Step 2: Eliminate [tex]\( y \)[/tex] by subtracting the equations
Next, we subtract equation (3) from equation (4) to eliminate [tex]\( y \)[/tex]:
[tex]\[
(24x + 2y) - (6x + 2y) = 44 - 8
\][/tex]
Simplifying the above equation:
[tex]\[
24x + 2y - 6x - 2y = 36
\][/tex]
[tex]\[
18x = 36
\][/tex]
### Step 3: Solve for [tex]\( x \)[/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[
x = \frac{36}{18} = 2
\][/tex]
### Step 4: Substitute [tex]\( x \)[/tex] back into one of the original equations to find [tex]\( y \)[/tex]
We substitute [tex]\( x = 2 \)[/tex] back into the first equation:
[tex]\[
6x + 2y = 8
\][/tex]
Substitute [tex]\( x = 2 \)[/tex]:
[tex]\[
6(2) + 2y = 8
\][/tex]
[tex]\[
12 + 2y = 8
\][/tex]
Solving for [tex]\( y \)[/tex]:
[tex]\[
2y = 8 - 12
\][/tex]
[tex]\[
2y = -4
\][/tex]
[tex]\[
y = \frac{-4}{2} = -2
\][/tex]
### Conclusion
The solution to the system of equations is:
[tex]\[
(x, y) = (2, -2)
\][/tex]
So the correct ordered pair is:
[tex]\[
\boxed{(2, -2)}
\][/tex]
Thus, the correct answer is:
[tex]\[
\text{D. } (2, -2)
\][/tex]
